The denominator (n-1) can be the product 7*17 = 119, because 12/17 = 84/119 and 5/7 = 85/119 , sosince 84 and 85 are 1 apart, we're there. Pat has 84 women classmates, and Chris has 85.Don't forget the 119 doesn't count the observer, so there are 120 students in the class.

Problem #22 - Posted Thursday, August 20, 1998

How Much Space?

A wire belt is wound tightly around the equator of the earth (circumference 25,000 miles), then 25 feet of wire is added and the belt is propped up at an equal height all the way around the planet. How much space will there be under the wire? (Please explain your reasoning.)

a) Not enough for an ant to crawl under b) Enough room for an ant, but not a mouse,

c) A Siamese cat can just squeeze under it d) Dan the math teacher could limbo under it!

Solution:

This problem is the same whether the original wire is wound around the earth, the moon, or an orange!

Let d = the space under the wire, and r = the radius, in feet, of the earth (or an orange, for that matter).The circumference of the earth is 2 Pi r feet, so the circumference of the wire is 2 Pi r + 25 feet.

The radius of the loop of wire is (2 π r + 25) / 2π = r + (25 / 2π), so the extra radius (the space under the wire) is 25 / 2π feet, which works out to about 25 / 6.2831853 ; approx. 3.9789 feet.

(even Dan the Math teacher should be able to get under that... wanna see?)

What is the smallest number that equals the sum of two (positive) perfect cubes in two different ways? (For example, 65 is the sum of two squares in two different ways: 65 = 8^2 + 1^2 = 7^2 + 4^2.)

Solution: This is the famous "Ramanujan taxicab number."

Prof. J.E. Littlewood, on a visit to Ramanujan at Trinity College in England, remarked that his taxicab number, 1729, was a rather dull number. "Oh, no," replied Ramanujan, "it is a very interesting number. It is the smallest number that is the sum of two cubes in two different ways!"

(dan's question: Does this contradict Fermat's Last Theorem?)

Problem #24 - Posted Saturday, September 12, 1998

That's Sum Product!

Pat and Chris are gambling in Las Vegas. I asked them to 'put their winnings together.'

Another way (thanx Trevor) is to plot the two curves x + √y = 7 and y + √x = 11 and see where they intersect.

Problem #26 - Posted Friday, October 9, 1998

The Two Elephants: Tons of Tens, or Tens of Tons?

Tenny and Tonny are two elephants. Every winter, Tenny gains ten percent of his body weight, and Tonny loses ten percent. Every summer, the opposite happens: Tenny loses ten percent and Tonny gains ten percent. This has gone on for ten years, and now they each weigh ten tons.

How many tons did Tenny and Tonny weigh ten years ago?

(explain steps, round to nearest pound.)

Solution:

Along my favorite theme: "Things are often not what they seem," the answer is NOT TEN TONS.

See, 90% of 110% of any number n is (.90)(1.10)(n) = 0.99 n. Same goes for 110% of 90%.

Problem #27 - Posted Wednesday, October 21, 1998

How Many Tiles?

A rectangular floor is composed of whole square tiles. A diagonal line is drawn and ruins some of the tiles. (See that on a 2 x 5 floor, 6 tiles are ruined, on a 2 x 4, only 4 are ruined.)

a) How many tiles are ruined on a 4 by 6 floor?

b) How about a 63 by 81 floor?

c) Generalize to an m by n floor.

Solution: (by Trevor Bird "as told to" Dan Bach)

Let's define a "tile function" T(m,n) = number of ruined tiles in an m by n rectangle.First let's look at the case where m < n have no common factors; every time the diagonalline crosses a vertical or horizontal line, a new tile is ruined, and there are m-1 verticaland n-1 horizontal lines; add the initial tile and get T(m,n) = m+n-1.

If m and n have a common factor g then there will be a pattern that repeats g times.The size of the repeating pattern will be m/g by n/g . If g is the greatest common factor(the GCD of m and n) then we have a total of T(m,n) = g(m/g + n/g - 1) ruined tiles.Another way to write this is T(m,n) = m + n - g .

(Andy M. also observed that if m goes into n evenly then there are n ruined tiles.)

Problem #28 - Posted Monday, November 2, 1998

Where On Earth?

I start running at 12:00, go 2 miles south by 12:15, 2 miles west by 12:30, and 2 miles north by 12:45. After 45 minutes I'm right back where I started!

Where on Earth am I? (there's more than one possibility)

Solution:(partly from new contestant Beth Wilson.)

The North Pole is the "top" answer, but there are an infinite number of circles near the South Pole where I could have started. One example is two miles north of any point at which the circumference of the circle that parallels the Equator is 2 miles (the radius would be 1/π) so that after running two miles you will be back where you started.

The other points follow this same pattern, except that the circumference of the circle is 2/2, 2/3, 2/4, 2/5, 2/6 ..., and so on!

Problem #29 - PostedFriday, November 13, 1998

Fridays the Thirteenth

What is the maximum number of Friday-the-13ths that there can be in a single year?

What is the minimum number?

Solution:

A regular year can start with Jan 1 on any of the 7 days of the week, and so can a leap year.So there are 14 possible "year-patterns"; these are the days for the 13th of each month:Notice the 13th is delayed the next month by the number of days above 28; 3 for Jan (31) etc.Fr Mo Mo Th Sa Tu Th Su We Fr Mo We. (reg) , and Fr Mo Tu Fr Su We Fr Mo Th Sa Tu Th (leap).

As Beth Wilson notes, the max number of times a day appears is 3, and the min is 1.A thorough check yields the fact that indeed in either a normal year or a leap year there area max of three Friday the 13ths and a min of one Friday the 13th.

(dan's incorrect note: i thought i remembered a year when there were no friday the 13ths!)